Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z^2 + z}{3z - 30} \times \dfrac{z^2 - 6z - 40}{z^3 - 2z^2 - 3z} $
Explanation: First factor out any common factors. $x = \dfrac{z(z + 1)}{3(z - 10)} \times \dfrac{z^2 - 6z - 40}{z(z^2 - 2z - 3)} $ Then factor the quadratic expressions. $x = \dfrac {z(z + 1)} {3(z - 10)} \times \dfrac {(z - 10)(z + 4)} {z(z + 1)(z - 3)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {z(z + 1) \times (z - 10)(z + 4) } {3(z - 10) \times z(z + 1)(z - 3) } $ $x = \dfrac {z(z - 10)(z + 4)(z + 1)} {3z(z + 1)(z - 3)(z - 10)} $ Notice that $(z + 1)$ and $(z - 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {z(z - 10)(z + 4)\cancel{(z + 1)}} {3z\cancel{(z + 1)}(z - 3)(z - 10)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $x = \dfrac {z\cancel{(z - 10)}(z + 4)\cancel{(z + 1)}} {3z\cancel{(z + 1)}(z - 3)\cancel{(z - 10)}} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $x = \dfrac {z(z + 4)} {3z(z - 3)} $ $ x = \dfrac{z + 4}{3(z - 3)}; z \neq -1; z \neq 10 $